1 Homework / Exam Reading –PAL, pp Homework –mp2 due before class number 12 Exam #1 –Class 13 (three sessions from today) –Open book / Open notes –Practice exam posted on web site now

2 Using C Structs in Assembly Code How do we access a C structure such as: #define NAMELEN 20 struct teststruct { int x, int y; char name[NAMELEN]; }t; t.x = 2; t.y = 5; strncpy(t.name, “wilson”, NAMELEN); trystruct(&t); /* pass to asm via pointer*/

3 Using C Structs in Assembly Code Assembly code would look like: movl4(%esp),%edx# ptr to t movl(%edx),%eax# x itself movl4(%edx),%ebx# y itself movb8(%edx),%cl# 1st string char 0xfffffc &t 0xfffff80xfffff4 %esp Return %eip t.x‘w’ ‘i’ ‘l’ ‘s’ ‘o’ ‘n’ ‘\0’t.y 0x0200e00x0200e40x0200e Stack struct teststruct

4 Using C Structs in Assembly Code However, we would normally have a pointer to string: #define NAMELEN 20 char array [NAMELEN]; struct teststruct { int x, int y; char *name; }t; t.x = 2; t.y = 5; t.name = array; strncpy(array, “wilson”, NAMELEN); trystruct(&t); /* pass to asm via pointer*/

5 Using C Structs in Assembly Code Assembly code would look like: movl4(%esp),%edx# ptr to t movl(%edx),%eax# x itself movl4(%edx),%ebx# y itself movl8(%edx),%edx# ptr to string movb(%edx),%cl # first string char 0xfffffc &t 0xfffff80xfffff4 %esp Return %eip t.x ‘w’ ‘i’ ‘l’ ‘s’ ‘o’ ‘n’ ‘\0’ t.y 0x0200e00x0200e40x0200e Stack struct teststruct t.name 0x

Introduction to Shift Instructions We can shift the bits in a byte, word, or long word by a variable number of positions These are the machine level instructions used to implement the C language operators > –SAL / SHL are the left shift instructions for signed or unsigned data (arithmetic or logical left shift) –SAR is the right shift instruction for signed data (arithmetic right shift) –SHR is the right shift instruction for unsigned data (logical right shift) 6

Introduction to Shift Instructions The SAL / SHL Instruction (Signed / Unsigned) The SAR Instruction (Signed) The SHR Instruction (Unsigned) 7 0 CF 0

Introduction to Shift Instructions The target of the shifting can be a register or memory location (byte, word, or long word) The count for the number of bits to shift can be specified with immediate data (constant) or the %cl register (variable) Examples: sall $4, %eax # logical left shift of %eax by 4 bits sarb %cl, label # arithmetic right shift of memory byte # by a variable value stored in the %cl 8

Introduction to Shift Instructions Multiplication by 2 N can be done via left shift sall $4, %eax # %eax times 2 4 Can combine left shifts and addition Division by 2 N can be done via right shift sarb %cl, label # memory byte / 2 %cl Can combine right shifts and subtraction 9

10 Introduction to Multiply and Divide Unsigned Multiply and Divide –mul –div Signed Multiply and Divide –imul –idiv We won’t do much with these because of the complexity involved - especially for divide

11 Introduction to Multiply and Divide Multiply always operates with %al, %ax, or %eax Result needs more bits than either operand Syntax: mulb %bl %ax  %al * %bl mulw %bx %dx, %ax  %ax * %bx mull %ebx %edx, %eax  %eax * %ebx

12 Introduction to Multiply and Divide Register Pictures (Byte) (Word) %bl %ax * %bx %dx, %ax *

13 Example – For/While Loop and mul C code for n = 5! (done as a for loop) unsigned int i, n; n = 1; for (i = 1; i <= 5; i++) n *= i; C code for n = 5! (done as a while loop) unsigned int i, n; n = i = 1; while (i <= 5) n *= i++;

14 Example – For/While Loop and mul Assembly code for n = 5! (byte * byte = word) movb $1, %bl # i = 1 movb %bl, %al # n = i = 1 loop:cmpb $5, %bl # while (%bl <= 5) jaexit # %bl > 5 now mulb%bl # %ax = %al * %bl incb%bl # incr %bl jmploop # and loop exit: # 5! in %ax Note: No difference between for and while in assy

15 Example – For/While Loop and mul Assembly code for n = 5! (word * word = long) movw $1, %bx # i = 1 movw %bx, %ax # n = i = 1 loop:cmpw $5, %bx # while (%bx <= 5) jaexit # %bx > 5 now mulw%bx # %ax = %ax * %bx # %dx = 0 now incw%bx # incr %bx jmploop # and loop exit: # 5! in %eax

16 Recursive Factorial Main program to call recursive factorial subr.text pushl $5 call factorial addl $4, %esp ret

17 Recursive Factorial factorial: # works up to 16 bit results movl 4(%esp), %eax cmpl $1, %eax jna return decl %eax pushl %eax call factorial addl $4, %esp movw 4(%esp), %bx mulw %bx # 16 lsbs go to %ax return: # ignore msbs in %dx ret.end

18 Recursive Factorial Stack Operations (while calling) Stack Operations (while returning) %esp at main value 5 %eip (main) 1 st Call value 4 %eip (fact) value 3 %eip (fact) value 2 %eip (fact) value 1 %eip (fact) Decr & 2 nd Call Decr & 3 rd Call Decr & 4 th Call Decr & 5 th Call arg ==1 so 1 st Return Multiply & 2 nd Return Multiply & 3 rd Return Multiply & 4 th Return Multiply & 5 th Return %eax = 1%eax = 2%eax = 6%eax = 24%eax = 120